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The vapour pressure of pure benzene at a...

The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5g when added to 39.0 g of benzene (molar mass 78 g `"mol"^(-1)`), vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance ?

Text Solution

Verified by Experts

For any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.
The various quantities known to us are as follows:
`p_(1)^(0)0.580 "bar" p=0.485 " bar" 0.845" M_(1)=78 g "mol"^(-1), w_(2)=0.5g, w_(1)= 39 g`
Substituting these values in equation (2.28), we get
`(0.850 "bar"-0.845 "bar")/(0.850"bar")=(0.5gxx78 "g mol"^(-1))/(M_(2)xx39g)`
Therefore `M_(2)=170 "g mol"^(-1)`
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